Find the slope of the curve x^2 + xy + y^2 = 7 at (1,2)
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x^2 + xy + y^2 = 7
The slope is the derivative at (1,2).
Let us diffirentiate with respect to x:
=(x^2 + xy + y^2)' =...
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To do this problem we first apply the product rule:
d ( xy) / dx = y + x*dy/dx
to find the derivative of x^2 + xy + y^2 = 7.
So the derivative of x^2 + xy + y^2 = 7 is
2x + x*dy/dx +y + 2y* dy/dx =0
Now taking dy/dx to one side,
=> dy/dx( x +2y ) = -x^2 -xy - y^2
=> dy/dx = (-x^2 -xy - y^2) / (x+2y)
At (1,2)
dy/dx = (-x^2 -xy - y^2) / (x+2y)
=>( - 1^2 - 1*2 - 2^2)/ (1+ 4)
=> (-2 -2 -4 )/5
=> -8/5
Therefore the required slope is -8/5
The slope of the curve ,x^2+xy+y^2 = 7 at (1,2) is got by diffrentiating both sides of the equation and evaluating dy/dx or y' at (1,2).
x^2+xy+y^2 = 7
(x^2+xy+y^2)' = (7)'
2x+y+y' +2y*y' = 0
y'(1+2y) = -(2x+y)
y' = -(2x+y)/(1+2y)at(1,2) = -(2*1+2)/(1+2*2)
y' = -4/5
Slope = y' = -4/5 at (1,2).
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